are drawn across parallel lines, running the whole length of the scale, from each division on one side, to the next higher on the other. The parallel lines divide the width of the scale into 10 equal parts. Since each oblique line, then, in crossing the scale, passes over one division of length, it is evident that, in passing one tenth across, (that is to the first parallel line,) it will pass over one tenth of a division of length; in passing two tenths across, (that is, to the second parallel line,) it will pass over two tenths of a division of length, and so on. The parallel lines are numbered at the end of the scale. To take off a distance, containing hundreds, then, as 234, we must place one foot of the dividers on the second division of the larger scale, and on the parallel line marked 4, and extend the other foot to the third oblique line. Decimals may evidently be taken off in a similar manner; the divisions of the larger scale being made units, and those of the smaller, tenths and hundredths.]

NOTE. The length of the two legs may be found by measuring them upon the same scale of equal parts from which the hypothenuse was taken.

PROBLEM X. To make a right angled triangle, the angles and one leg being given. Fig. 31.

Fig. 31.

B

15

285

Suppose the angle at C 33° 15', and the leg A C 285. Draw the leg A C, making it in length 285; at A erect a perpendicular an indefinite length; at C make an angle of 33° 15'; through where that number of degrees cuts the arc, draw a line till it meets the perpendicular at B.

NOTE. If the given line C A should not be so long as the chord of 60°, it may be continued beyond A, for the purpose of making the angle.

PROBLEM XI. To make a right angled triangle, the hypothenuse and one leg being given. Fig. 32.

Fig. 32.

Suppose the hypothenuse A C 40, and the leg A B 28. Draw the leg A B in length 28; from B erect a perpendicular an indefinite length; take 40 in the dividers, and setting one foot in A, wherever the other foot strikes the dicular will be the point C.

perpen-.

NOTE. When the triangle is constructed, the angles may be measured by a protractor, or by a scale of chords.

PROBLEM XII. To make a right

angled triangle, the two legs being given.

Fig. 33.

Fig. 33.

AL

38

Suppose the leg A B 38, and the leg B C 46.

Draw the leg A B in length 38; from B erect a perpendicular to C, in length 46; and draw a line from A to C.

Suppose the side B C 98; the angle at B 45° 15', the angle at D 108° 30', consequently the other angle 26° 15'.

Draw the side B C in length 98; on the point B make an angle of 45° 15'; on the point C make an angle of 26° 15', and draw the lines B D and C D.

PROBLEM XIV. To make an oblique angled triangle, two sides and an angle opposite to one of them being given. Fig. 35.

Fig. 35.

Suppose the side B C 160, the side B D 79, and the angle at C 29° 9'.

Draw the side B C in length 160; at C make an angle of 29° 9′, and draw an indefinite line through where the degrees cut the arc; take 79 in the dividers, and with one foot in B lay the other on the line C D; the point D will be the other angle of the triangle.

PROBLEM XV.

Fig. 36.

To make an D

oblique angled triangle, two sides

and their contained angle being given. Fig. 36.

209

Suppose the side B C 109, the side B D 76, and the angle at B 101° 30'.

Draw the side B C in length 109; at B make an angle of 101° 30', and draw the side B D in length 76; draw a line from D to C, and it is done.

Draw the line A B the length of the proposed square; from B erect a perpendicular to C, and make it of the same length as A B; from A and C, with the same distance in the dividers, describe arcs intersecting each other at D, and draw the lines A D and D C.

Draw the line A B equal to the longest side of the rectangle; on B erect a perpendicular the length of the shortest side to C; from C, with the longest side, and from A, with the shortest side, describe arcs intersecting each other at D, and

PROBLEM XVIII. To describe a circle which shall pass through any three given points, not lying in a right line, as A, B, D. Fig. 39.

B

*

Fig. 38.

A

Draw lines from A to B, and from B to D; bisect those lines by PROBLEM II, and the point where the bisecting lines intersect each other, as at C, will be the centre of the circle.

PROBLEM XIX. To find the centre of a circle.

By the last PROBLEM it is plain, that if three points be any where taken in the given circle's periphery, the centre of the circle may be found as there taught.

Directions for constructing irregular figures of four or more sides, may be found in the following treatise on SUR

VEYING.

TRIGONOMETRY.

TRIGONOMETRY is that part of practical GEOMETRY, by which the sides and angles of triangles are measured whereby three things being given, either all sides, or sides and angles, a fourth may be found; either by measuring with a scale and dividers, according to the PROBLEMS IN GEOMETRY, or more accurately by calculation with logarithms, or with natural sines.

TRIGONOMETRY is divided into two parts, rectangular and oblique-angular.

PART I.

RECTANGULAR TRIGONOMETRY.

This is founded on the following methods of applying a circle to a triangle.

PROPOSITION I. In every right angled triangle, as A B C, Fig. 40, it is plain from Fig. 7, compared with the Geometrical definitions to which that Figure refers, that if the hypothe-A nuse A C be made radius, and with it an arc of a circle be described from each end, B C will be the sine of the angle at A, and A B the sine of the angle at C; that is, the legs will be sines of their opposite angles.

PROPOSITION II. If one leg, A B, Fig. 41, be made radius, and with it on the point A an arc be described, then B C, the other leg, will be the tangent, and A C the secant of the angle at A; and if B C be made radius, and an arc be described with it on the point C, then A B will be the tangent, and A C the se

Fig. 40.

C

Fig. 41.

B